it suffices to bound the latter factor. By the area formula [1]}, we have for any \(B \subset \mathcal {B}_1^d\)  that
\(\phi _{\star }\mu _d(B) &= \int _{\phi ^{-1}(B)} \mathrm {d}\mu _d(x) \\&= \int _{\phi ^{-1}(B)} \frac{\left| \det (\mathrm {D}\phi (x)) \right|}{\left| \det (\mathrm {D}\phi (x)) \right|} \mathrm {d}\mu _d(x) \\&\le \frac{1}{c} \int _{\phi ^{-1}(B)} \left| \det (\mathrm {D}\phi (x)) \right| \mathrm {d}\mu _d(x) \\&= \frac{1}{c} \int _B \left| \phi ^{-1}(y) \right| \mathrm {d}\mu _d(y) \le \frac{N}{c} \mu _d(y)\) 
